Power supply and low/high-level voltage

Community General Discussions
1

Hi everybody,

I am trying to have these two components work together:

EFM32GG980 : http://cdn.energymicro.com/dl/devices/pdf/d0045_efm32gg980_datasheet.pdf

Fastrax UC430 : http://www.fastraxgps.com/showfile.cfm?guid=fd5516a9-3c5d-4e30-8dc7-cbbb1b9102fa

 

The electrical specs are the following:

EFM32:

VDD: 1.8V-3.0V

Input Low/High: 0.3VDD / 0.7VDD

Output Low/High with [email protected]: 0.25VDD / 0.7VDD

Output Low/High with VDD@3V: 0.05VDD / 0.9VDD

UC430:

VDD: 1.8V

Input Low/High: 0.45V / 0.7VDD (max 3.6V)

Output Low/High: 0.4V / 0.75VDD

 

My question is quite simple. In your opinion, what would be the best way to have these work correctly together (power supply + logic levels)? This system will run on a little lithium battery, and my main requirement is to minimize the power consumption as much as possible. I would be grateful if you could give me your input on this.

Thanks a lot in advance for your help!

 

Cheers,

Chris

2

How much time will the

How much time will the microcontroller and GPS units spend active?
Are you looking to log a GPS sample once a minute, or will you be processing a bit more than that?
What other parts (if any) will be powered by the battery, or via the microcontroller?

If the percentage of time spent ‘awake’ will be fairly low, you can probably get away with a common linear voltage regulator, or possibly one regulator each for the microcontroller and GPS.

If the micro and GPS will be reasonably active, use a proper DC-DC buck converter to step down the voltage from your lithium battery.
Set the output to 1.87V - you’ll want an adjustable model so you can get the voltage between the minimum for the EFM32 and the maximum for the UC430. Alternatively you can raise the voltage output of the DC-DC converter, connect it directly to the EFM32, and use a Zener diode or similar to drop the voltage powering the UC430.

The best way to save power will be to make good use of the low power and hibernate modes offered by the EFM32 and UC430 respectively.

You’ll probably be ok as far as logic levels go, the only one I can see being a problem is the low level output from the UC430… if it’s close to the maximum of 0.4V the EFM32 may not recognise it. Do some testing and see what the actual low level output is, hopefully you can get away without requiring any extra components for level shifting.

3

Hi TeleFox,Thanks a lot for

Hi TeleFox,

Thanks a lot for taking time to help me!

I came more or less to the same conclusions as you. About your question, the system will spend most of its time in hibernation and will wake up only about once a minute. However, I guess I have no other choice than supplying power to the uC and GPS to keep them in hibernate mode. Therefore, I will need at least a way of supplying regulated current to the GPS, which, unlike the EFM, allows only for an operating supply voltage of 1.71V-1.89V. So, as you mentionned, I will need at least one LDO or buck converter in the system.

As a consequence, the question left to be answered is whether to use a LDO or buck converter. As said, the system will spend most of its time in hibernate mode, so with a current consumption of about 25uA. When active during short periods (a couple of seconds max), it will require maybe up to 50mA. As a consequence, what do you think as the most appropriate solution?

Considering that the converter would have to run continuously to supply power to the hibernating GPS and uC, I was maybe thinking about a buck converter because of the better efficiency (like the LTC3388-1), but I have very little experience with such design issues. What is your opinion?

Also, about the logic levels:

You’ll probably be ok as far as logic levels go, the only one I can see being a problem is the low level output from the UC430… if it’s close to the maximum of 0.4V the EFM32 may not recognise it. Do some testing and see what the actual low level output is, hopefully you can get away without requiring any extra components for level shifting.

Actually I miscalculated, you’re right, it should be fine. I don’t have the UC430 yet, so I haven’t been able to test this so far. In fact, by examining the specs again, I am not even worried about the logic low of the UC430, as it will output something under 0.4V and the EFM32 with will await something under 0.6V (0.3VDD). Am I wrong?

Thanks a lot in advance for your help!

Cheers,
Chris

4

Well the first thing to

Well the first thing to mention is that your buck converter will probably take up more space than a linear regulator, but that is only an issue if you’re really tight on room.

To compare the two options, we can set up a quick set of calculations.
If we estimate the ‘on’ time at 5s per minute in the worst case, then we’ll be consuming 50mA for 5s and ~25μA for 55s out of every minute.

The linear regulator will simply ‘burn off’ whatever voltage difference there is between the input and output, so in this case we’re losing 3.7V - 1.8V = 1.9V (assuming nominal 3.7V from your lithium battery).
Power lost in the regulator when the system is ‘on’ is: 1.9V x 50mA = 95mW
Energy lost in the regulator when the system is ‘on’ is: 95mW x 5s = 475mJ
Power lost in the regulator when the system is ‘off’ is: 1.9V x 25μA = ~0.05mW
Energy lost in the regulator when the system is ‘off’ is: 0.05mW x 55s = 2.75mJ

The total energy lost in the regulator every minute is therefore 475mJ + 2.75mJ = 477.75mJ

If we take the datasheet values for the LTC3388-1, we can see that it is ~90% efficient when supplied at 3V and producing 50mA, and ~85% efficient when producing 25μA. I’m going to ignore the ‘no load’ current draw since it’s less than 1μA.
Power lost in the converter when the system is ‘on’ is: 3.7V x 50mA x ((100%/90%) - 1) = 20.56mW
Energy lost in the converter when the system is ‘on’ is: 20.56mW x 5s = 102.7mJ
Power lost in the converter when the system is ‘off’ is: 3.7V x 25μA x ((100%/85%) - 1) = 0.02mW
Energy lost in the converter when the system is ‘off’ is: 0.02mW x 55s = 1.1mJ

The total energy lost in the converter every minute is therefore 102.7mJ + 1.1mJ = 103.8mJ

So in this case the converter is going to perform a lot better than the linear regulator. When you consider that the system will only consume around 450mJ per minute by itself, the improved efficiency of the buck converter really starts to look good.

5

Hi TeleFox,Thanks a lot for

Hi TeleFox,

Thanks a lot for your reply. It clearly answered my question.

Best regards,
Chris